Conditioning on Completed Sigma-Algebras

Why Completion Matters for Conditioning

A conditional expectation is specified by measurability and integral identities, but it is determined only up to almost-sure equality. This distinction matters when the conditioning sigma-algebra is completed: completion can make additional representatives measurable, even though it does not change the almost-sure information used to define the conditional expectation.

The key conclusion is that conditioning on a sigma-algebra and on its probability completion gives the same conditional expectation almost surely. The measurable functions that represent that conditional expectation need not be exactly the same at every point.

Relative Completion and Conditional Expectation

Definition: Let $(\Omega,\mathcal F,P)$ be a probability space and let $\mathcal G\subseteq\mathcal F$ be a sigma-algebra. The relative $P$-completion of $\mathcal G$ within $\mathcal F$, denoted by $\mathcal G^P$, consists of the sets $A\in\mathcal F$ for which there is a set $B\in\mathcal G$ such that $P(A\triangle B)=0$. In symbols, $$ \mathcal G^P = \{A\in\mathcal F:\text{ there is }B\in\mathcal G \text{ with }P(A\triangle B)=0\}. $$

Here $A\triangle B$ is the symmetric difference: the points belonging to exactly one of $A$ and $B$. Thus $\mathcal G^P$ contains every event in $\mathcal G$, as well as ambient measurable events that differ from such an event only on a null set. In particular, every $F$-measurable null set belongs to $\mathcal G^P$, by taking $B=\varnothing$.

This is a relative completion: it adds only sets that are already in $\mathcal F$. If $\mathcal F$ is complete, it includes every subset of every ambient null set. The definition gives a sigma-algebra, since complements preserve symmetric differences and countable unions of almost-surely equal events are almost-surely equal.

Definition: For an integrable random variable $X$, a conditional expectation of $X$ given $\mathcal G$ is an integrable, $\mathcal G$-measurable random variable $Y$ such that, for every $B\in\mathcal G$, $$ \int_B Y\,dP=\int_B X\,dP. $$ It is denoted by $\mathbb E[X\mid\mathcal G]$, with the understanding that it is defined up to almost-sure equality.

The integral identities are insensitive to changes on null sets. Measurability is the part that can change when we replace $\mathcal G$ by $\mathcal G^P$.

Completion Can Add New Representatives

Worked Example: A Conditional Expectation with Two Measurable Versions

Let $\Omega=\{a,b\}$, let $\mathcal F$ be the power set of $\Omega$, and set $P(\{a\})=1$, $P(\{b\})=0$. Take the trivial sigma-algebra $\mathcal G=\{\varnothing,\Omega\}$ and $X=1_{\{b\}}$. Since $X=0$ almost surely, the constant random variable $Y=0$ is a conditional expectation given $\mathcal G$: it is $\mathcal G$-measurable, and for both events in $\mathcal G$ its integral equals the integral of $X$.

The completion $\mathcal G^P$ is all of $\mathcal F$. Indeed, $\{b\}$ differs from $\varnothing$ by a null set, and $\{a\}$ differs from $\Omega$ by a null set. Now $X$ itself is $\mathcal G^P$-measurable and is also a conditional expectation given $\mathcal G^P$: for every $A\in\mathcal G^P$, $\int_A X\,dP=\int_A X\,dP$. The versions $0$ and $X$ differ at $b$, but only on a null set.

This example illustrates the precise role of completion. The completed sigma-algebra permits a version to be changed on an ambient null set while remaining measurable. Such a change need not preserve pointwise values, but it preserves the almost-sure equivalence class.

A Measurable-Version Lemma

We will use the fact that a real-valued random variable measurable with respect to $\mathcal G^P$ has a $\mathcal G$-measurable version. Here, a version means a random variable equal to the original almost surely. The statement explains why completion adds representatives rather than new almost-sure classes of real-valued random variables.

Lemma: If $Z$ is a finite, real-valued, $\mathcal G^P$-measurable random variable, then there is a finite, real-valued, $\mathcal G$-measurable random variable $Z_0$ such that $Z_0=Z$ almost surely.

Proof. For each $n\in\mathbb N$, truncate $Z$ to the interval $[-n,n]$ and round down to the nearest multiple of $2^{-n}$. Call the resulting finite-valued random variable $S_n$. It is $\mathcal G^P$-measurable, and $S_n\to Z$ pointwise because $Z$ is finite.

Write the finitely many level sets of $S_n$ as $C_{n,1},\ldots,C_{n,m_n}$, all in $\mathcal G^P$. By the definition of $\mathcal G^P$, for each $C_{n,j}$ there is $D_{n,j}\in\mathcal G$ with $P(C_{n,j}\triangle D_{n,j})=0$. Assign to $D_{n,j}$ the value of $S_n$ on $C_{n,j}$, and sum these assigned values times $1_{D_{n,j}}$, obtaining a finite-valued $\mathcal G$-measurable random variable $T_n$. The sets $D_{n,j}$ need not form a partition everywhere, but outside the finite union of the symmetric differences they agree with the level sets of $S_n$; hence $T_n=S_n$ there.

The union of these exceptional sets over all $n$ is null, since it is a countable union of null sets. Thus $T_n=S_n$ eventually at every point outside a single null set, and there $T_n\to Z$. The set on which the sequence $(T_n)$ converges is $\mathcal G$-measurable, by the Cauchy criterion expressed using countably many inequalities. Define $Z_0$ to be its limit on that set and zero elsewhere. Then $Z_0$ is finite and $\mathcal G$-measurable, and $Z_0=Z$ outside the exceptional null set. $\square$

Conditional Expectations Are Unchanged Almost Surely

Theorem: Let $X$ be integrable on $(\Omega,\mathcal F,P)$, and let $\mathcal G\subseteq\mathcal F$. Then a conditional expectation of $X$ given $\mathcal G$ is also a conditional expectation of $X$ given $\mathcal G^P$. Consequently, $$ \mathbb E[X\mid\mathcal G^P] = \mathbb E[X\mid\mathcal G] \quad\text{almost surely}. $$

Proof. Let $Y=\mathbb E[X\mid\mathcal G]$. Since $\mathcal G\subseteq\mathcal G^P$, $Y$ is $\mathcal G^P$-measurable. Take any $A\in\mathcal G^P$. By definition there is $B\in\mathcal G$ with $P(A\triangle B)=0$. Integrable random variables have the same integral over two events that differ by a null set. Therefore, $$ \int_A Y\,dP = \int_B Y\,dP = \int_B X\,dP = \int_A X\,dP. $$ The middle equality is the defining property of $Y$, since $B\in\mathcal G$. Thus $Y$ satisfies the defining properties of a conditional expectation given $\mathcal G^P$.

For completeness, uniqueness up to almost-sure equality follows directly from the defining integral identities. If $U$ and $V$ are both conditional expectations given $\mathcal G^P$, then $D=U-V$ is integrable and $\mathcal G^P$-measurable, and $\int_A D\,dP=0$ for every $A\in\mathcal G^P$. In particular, take $A=\{D>0\}$. If this event had positive probability, then $\{D>0\}$ would be the union over $n\in\mathbb N$ of $\{D\ge 1/n\}$, so one such set would have positive probability and $\int_{\{D>0\}}D\,dP>0$, a contradiction. Applying the same argument to $-D$ shows $D=0$ almost surely. Hence the conditional expectations given $\mathcal G$ and $\mathcal G^P$ agree almost surely. $\square$

Checking the Integral Identities in a Partition Example

Worked Example: Conditioning on Two Half-Intervals

Let $\Omega=[0,1]$ with Lebesgue probability measure, and let $\mathcal G$ be generated by $H=[0,1/2]$. This sigma-algebra has four events: $\varnothing$, $H$, $H^c$, and $\Omega$. Let $X=1_{[0,1/4]}$. Define $$ Y=\tfrac12 1_H. $$

The variable $Y$ is $\mathcal G$-measurable. On $H$, the integrals are $\int_H Y\,dP=(1/2)(1/2)=1/4$ and $\int_H X\,dP=1/4$. On $H^c$, both integrals are zero. The identities for $\varnothing$ and $\Omega$ follow by adding these values, so $Y$ is a conditional expectation of $X$ given $\mathcal G$.

Now let $A\in\mathcal G^P$. It differs by a null set from some $B\in\mathcal G$. The integrals of $X$ and $Y$ over $A$ therefore equal their respective integrals over $B$, which were just checked to be equal. This verifies directly that the same $Y$ is a conditional expectation given the completed sigma-algebra.

Worked Example: Changing a Version at a Null Point

Continue with Lebesgue measure on $[0,1]$ and $\mathcal G=\sigma([0,1/2])$. The singleton $\{0\}$ is ambient-measurable and null, so $\{0\}\in\mathcal G^P$, although $\{0\}\notin\mathcal G$. With $X=1_{[0,1/4]}$ and $Y=\tfrac12 1_{[0,1/2]}$ as above, define $Y'=Y+1_{\{0\}}$.

The random variable $Y'$ is $\mathcal G^P$-measurable and differs from $Y$ only at $0$. For every $A\in\mathcal G^P$, the integral of $Y'$ over $A$ equals the integral of $Y$ over $A$, because $\int_A1_{\{0\}}\,dP=0$. Since $Y$ already satisfies the conditional-expectation identity for every such $A$, $Y'$ does too. Completion allows this additional measurable version, but does not alter the almost-sure conditional expectation.

What Completion Does—and Does Not Do

Completion does not provide additional information about $X$ in the sense of changing its conditional expectation as an almost-sure class. Every completed event agrees almost surely with an original conditioning event, and the defining integrals cannot distinguish between events that agree almost surely.

It does affect the available pointwise representatives. A function obtained by modifying a $\mathcal G$-measurable version on a null set may be $\mathcal G^P$-measurable even when it is not $\mathcal G$-measurable. The version lemma gives the converse perspective: any finite real-valued $\mathcal G^P$-measurable random variable has a $\mathcal G$-measurable version equal to it almost surely.

A common pitfall is to conclude that the two conditional expectations must be equal at every point. The defining uniqueness is only almost sure. Pointwise equality can depend on which measurable versions are selected. When a calculation or later argument needs a particular pointwise version, its measurability and its values on null sets should be checked separately.

Check Your Understanding

Use the definitions and integral criterion to answer the following questions.

Why does every ambient-measurable null set belong to $\mathcal G^P$?
What two properties must a random variable satisfy to be a conditional expectation given $\mathcal G$?
In the theorem’s proof, why can the integral over $A\in\mathcal G^P$ be replaced by the integral over $B\in\mathcal G$?
How can a conditional expectation given $\mathcal G^P$ differ pointwise from a $\mathcal G$-measurable version?
Does completion change the almost-sure equivalence class of the conditional expectation? Explain briefly.